# Does $\alpha$ need to be transcendental over F?

In the book there is this exercise:

Let E be an extension fiel of F, with $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over F but algebraic over $F(\beta)$. Show that $\beta$ is algebraic over $F(\alpha)$.

The suggested solution was:

It doesn't seem that they use that $\alpha$ is transcendental in F, or do they? Or is the statement valid without this assumption?

• There is a polynomial P($\alpha$) = 0 with coefficients polynomials f($\beta$). You do just rewrite this polynomial to conclude. – Piquito Apr 23 '15 at 18:17

a polynomial in $\alpha$ with coefficients that are polynomials in $\beta$ can be formally rewritten as a polynomial in $\beta$ with coefficients that are polynomials in $\alpha$
to conclude that this new polynomial (seen in $(F[\alpha])[x]$) is not the zero polynomial. This is necessary to conclude that $\beta$ satisfy a non-zero polynomial over $F(\alpha)$ so $\beta$ is algebraic over $F(\alpha)$.
• @DanielEscudero: doing so might be difficult. You'd have to ensure the minimal polynomial of $\alpha$ over $F$ divides all the coefficients of the polynomial obtained by rewriting as polynomial in $\beta$. I can't see of any easy way to rig this. Roots of unity might be a good place to start, but that's just an offhand guess. – Alex Wertheim Apr 23 '15 at 18:12
• Is the point that when we reqrite the poluynomial we have to make sure that not coefficients infront of each $\beta$ is zero? So we are using that $\alpha$ is transcendental in F, on each coefficient, not on the polynomial as a whole? – user119615 Apr 23 '15 at 18:25